### Video Transcript

And this trouble we have been given that a queen ‘s task three times. So get ‘s first record the sample space. So the sample space here will be let ‘s barely write that. So we will precisely repeat the sample distance for The head, the queen just for two times. And then we will convert that To generate the samples piece for mint job for three times by fair adding edge in the inaugural row and tea on the second road. So that ‘s a simple promptly besides right down the sample distribution space far all the outcomes Of the coins when it is tasked three times. So we need to determine the probability Of precisely three heads. So we observed that there is only one prison term the three all the three heads can be occurring when we just trust the coin three times the probability of all three heads. That is one ten 8. And now we need to determine Probability of precisely one question. So one head, we can observe the sample space that is contented in this. It ‘s contained in this equally well as this. So we can see that one head can occur three times out of eight times. So that ‘s the necessitate probability. And now in the next case we are given that at least one promontory is observed and we need to determine the probability That at least two heads turn up. thus hera we will consider the consequence millimeter as that you would win At least one lead is observed. indeed in that font it ‘s probability will be seven x 8 because at least one had will count all the outcomes other than this tail tail aim, So that ‘s seven x 8 and the probability went at least two heads can be seen. So let ‘s take that probability belong to even be. So we good need to mark down those Outcomes which has at least two heads. So this is the one, this has at least two heads, this has at least two heads And this has at least two hits. So that comes out to before over eight. So that ‘s basically four by eight and we need to determine the probability of the event B provided that he has already happened. so in that casing this will be obtained by taking the probability of a intersection of these two events divided by the probability of A. sol when we see the intersection of even A. And even to be. So that comes out to be probability of evening be only because this is included in all the most of their currents of even. So we can say this is probability of B divided by probability of And that comes out to before by eight, divided by seven x 8. so when we simplify we can get the necessitate probability as 4,57

## Leave a Comment